How to embed random walk into Brownian Motion (Donsker's Invariance Principle)?

Question

I am trying to understand the proof of Donsker's invariance principle:

First some definitions Let $X_1,X_2,...$ be i.i.d. real-valued random variables with mean 0 and variance 1. We define $S_0=0$ and $S_n= X_1+ ... + X_n$ for $n \geq 1$. To get a process in continuous time, we interpolate linearly and define for all $t \geq 0$ $$ S_t = S_{[t]}+ (t-[t])(S_{[t]+1}- S_{[t]}). $$ Then we define for all $t \geq 0$ $$ S^*_n(t)= \frac{S_{nt}}{n}. $$ Let $C([0,1])$ be the space of real-valued continuous function defined on $[0,1]$ and endow space with the supremumnorm. Then $(S^*_n(t))_{0 \leq t \leq 1}$ can be seen as a random variable taking values in $C([0,1])$. Now let $\mu_n$ be its law on that space of continuous functions and let $\mu$ be the law of Brownian motion on $C([0,1])$. Then the following holds:

Theorem (Donsker): The probability measure $\mu_n$ converges weakly to $\mu$, i.e. for every $F: C([0,1]) \rightarrow \mathbb{R}$ bounded and continuous, $$ \int F d\mu_n \rightarrow \int F d\mu $$ as $n \rightarrow \infty$.

The proof I saw in class, is very similar in spirit to the one that can be found here (Theorem 12, page 23). The idea is to start with a given Probability space $(\Omega, \mathcal{F},P)$ where a Brownian Motion $(B_t)_{t \geq 0}$ already exists and then embedd a random walk into the Brownian Motion.

To do that, let $(\mathcal{F}_t)_{t \geq 0}$ be the filtration generated by Brownian motion, i.e. $\mathcal{F}_t := \sigma(B_s: 0 \leq s \leq t)$. Then we define $\tau_0 := 0$ and for $i \geq 0$ we define inductively $$ \tau_{i+1}:= \inf \{ t > \tau_i : |B_t - B_{\tau_i}| = 1\} $$ I could show that $(\tau_i)_{i \geq 0}$ is an increasing sequence of stopping times. And due to the fact that for a Brownian Motion we have $\limsup_{t \rightarrow \infty}B_t= \infty$ almost surely, there exists a set of probability 1 where for every $i \geq 0$ the stopping time $\tau_i$ is finite.

Then it is claimed that due to the strong Markov property applied at time $\tau_i$, the random vector $(\tau_{i+1}-\tau_i, B_{\tau_i +1}- B_{\tau_i})$ is independent of $\mathcal{F}_{\tau_i}$ (the stopped sigma-algebra at $\tau_i$) and distributed as $(\tau_1,B_{\tau_1})$. This is the part I do not understand. The strong Markov property I know of is the following:

Theorem (Strong Markov property) Let $T$ be a stopping time such that $P[T < \infty] > 0$. For every $t \geq 0$ we put $$ B_t^{(T)} = \mathbb{1}_{\{T < \infty\}}(B_{t+T}-B_T). $$ Then under $P[\cdot | T < \infty]$ the process $B^{(T)}$ is a Brownian Motion independent of $\mathcal{F}_T$

So the questions I have are the following:

1. Why is $(\tau_{i+1}-\tau_i, B_{\tau_i +1}- B_{\tau_i})$ independent of $\mathcal{F}_{\tau_i}$ ?
2. Why is $(\tau_{i+1}-\tau_i, B_{\tau_i +1}- B_{\tau_i})$ distributed as $(\tau_1, B_{\tau_1})$ ?

Sorry for the long post, I wanted to include all the details. Thanks a lot in advance!

Answer 1

I suppose it is a bit late for you now, but I suppose it could still be useful for others, so I will answer anyway.

Due to the Strong Markov Property and the recurrence of Brownian Motion, $B^i_t=B^{(\tau_i)}_t:= B_{\tau_i+t}-B_{\tau_i} $ is a Brownian motion independent of $\mathcal{F}_{\tau_i}$.

Define $\tau^*_i := \inf\{t \ge 0: |B^{(\tau_i)}_t|\ge 1\}$. It is clear that $B^{(\tau_i)}_{\tau^*_i} = B_{\tau_{i+1}}-B_{\tau_{i}}$ pointwise. However, as $B^{(\tau_i)}$ is independent from $\mathcal{F}_{\tau_i}$ and $\tau^*_i$ is a stopping time for it, we get $[\tau^*_i \le t] \in \sigma (B^{(\tau_i)}_s: s \in [0,t])\perp \mathcal{F}_{\tau_i}$, hence $\tau^*_i = \tau_{i+1}-\tau_i$ is independent from $\mathcal{F}_{\tau_i}$. Now, you can show that $$ \Big[\tau^*_i \in [s_0,s_1], B^{(\tau_i)}_{\tau^*_i} \in [a,b]\Big] \in \sigma\Big( B^{(\tau_i)}_s, s \le s_1 \Big) \perp \mathcal{F}_{\tau_i}. $$

Which answers the first question. For the second, just notice again the $\tau^*_i= \tau_{i+1}-\tau_i$. However, as both $B^{(\tau_i)}_t$ and $B_t$ are Brownian motions, therefore, $\tau_1$ and $\tau^*_i$ have the same distribution. The question then follow again by using that both $B_t$ and $B^{(\tau_i)}_t$ are Brownian Motions.